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SCIENCE CHINA Information Sciences, Volume 63 , Issue 12 : 222303(2020) https://doi.org/10.1007/s11432-020-3033-3

Statistical CSI based design for intelligent reflecting surface assisted MISO systems

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  • ReceivedJun 14, 2020
  • AcceptedAug 12, 2020
  • PublishedOct 30, 2020

Abstract


Acknowledgment

This work was supported by National Key RD Program of China (Grant No. 2019YFB1803400), the NSFC-Zhejiang Joint Fund for the Integration of Industrialization and Informatization (Grant No. U1709219), and National Natural Science Foundation of China (Grant No. 61922071).


Supplement

Appendix

Proof of Proposition 3.1

Applying the Jensen's inequality, we have \begin{align}{\rm E}\left\{ \log_2 \left( 1+\gamma \right) \right\} \lessapprox \log_2 \left(1+{\rm E}\{\gamma\} \right) = \log_2 \left( 1+\gamma_0 {\rm E} \left\{ \left|({\boldsymbol h}_2^{\rm T} \boldsymbol{\Phi} {\boldsymbol H}_1+\lambda{\boldsymbol g}^{\rm T}) {\boldsymbol f} \right|^2 \right\} \right). \tag{29} \end{align} The remain task is the derivation of ${\rm~E}~\{~\left|({\boldsymbol~h}_2~\boldsymbol{\Phi}~{\boldsymbol~H}_1+\lambda{\boldsymbol~g})~{\boldsymbol~f}~\right|^2~\}$. Applying the binomial expansion theorem, we have \begin{align}& \left| \left({\boldsymbol h}_2^{\rm T} \boldsymbol{\Phi} {\boldsymbol H}_1+\lambda {\boldsymbol g}^{\rm T} \right) {\boldsymbol f} \right|^2 \tag{30} \\ &= \left| \left(a_2 {\bar{\boldsymbol h}}_2^{\rm T}+ b_2 {\tilde{\boldsymbol h}}_2^{\rm T} \right) \boldsymbol{\Phi} \left(a_1 {\bar{\boldsymbol H}}_1 + b_1 {\tilde{\boldsymbol H}}_1 \right) {\boldsymbol f} + \lambda\left(a_0 {\bar{\boldsymbol g}}^{\rm T} + b_0 {\tilde{\boldsymbol g}}^{\rm T} \right) {\boldsymbol f} \right|^2 \\ &= \left| x_1 + x_2 + x_3 + x_4 + x_5 \right|^2, \end{align} where \begin{align}& x_1 = (a_2 a_1 {\bar{\boldsymbol h}}_2^{\rm T} \boldsymbol{\Phi} {\bar{\boldsymbol H}}_1+ \lambda a_0 {\bar{\boldsymbol g}}^{\rm T}) {\boldsymbol f}, \tag{31} \\ & x_2 = a_2 b_1 {\bar{\boldsymbol h}}_2^{\rm T} \boldsymbol{\Phi} {\tilde{\boldsymbol H}}_1 {\boldsymbol f}, \tag{32} \\ & x_3 = b_2 a_1 {\tilde{\boldsymbol h}}_2^{\rm T} \boldsymbol{\Phi} {\bar{\boldsymbol H}}_1 {\boldsymbol f}, \tag{33} \\ & x_4 = b_2 b_1 {\tilde{\boldsymbol h}}_2^{\rm T} \boldsymbol{\Phi} {\tilde{\boldsymbol H}}_1 {\boldsymbol f}, \tag{34} \\ & x_5 = \lambda b_0 {\tilde{\boldsymbol g}}^{\rm T} {\boldsymbol f}. \tag{35} \end{align} It is easy to observe that $x_1$ is a constant and ${\rm~E}\{x_i\}~=~0$ holds for $i~=~2,3,4,5$. Besides, since ${\tilde{\boldsymbol~h}}_2$, ${\tilde{\boldsymbol~H}}_1$ and ${\tilde{\boldsymbol~g}}$ have zero means and are independent with each other, we can derive that \begin{align}& {\rm E} \left\{ \left|({\boldsymbol h}_2^{\rm T} \boldsymbol{\Phi} {\boldsymbol H}_1+\lambda {\boldsymbol g}^{\rm T}) {\boldsymbol f} \right|^2 \right\} \tag{36} \\ & = {\rm E} \{ | x_1 + x_2 + x_3 + x_4 + x_5 |^2 \} \\ & = |x_1|^2 + {\rm E}\{|x_2|^2\} + {\rm E}\{|x_3|^2\} + {\rm E}\{|x_4|^2\} + {\rm E}\{|x_5|^2\}. \end{align} Let ${\boldsymbol~w}~\triangleq\tilde{\boldsymbol~H}_1~{\boldsymbol~f}$. ${\rm~E}\{|x_2|^2\}~\in~\mathbb{C}^{N~\times~1}$ can be expressed as \begin{align}{\rm E}\{|x_2|^2\}= a_2^2 b_1^2 {\rm E} \left\{ \mathsf{tr} \left( \bar{\boldsymbol h}_{2}^{*} \bar{\boldsymbol h}_{2}^{\rm T} {\boldsymbol w} {\boldsymbol w}^{\rm H} \right) \right\}=a_2^2 b_1^2 \mathsf{tr} \left( \bar{\boldsymbol h}_{2}^{*} \bar{\boldsymbol h}_{2}^{\rm T} {\rm E} \left\{ {\boldsymbol w} {\boldsymbol w}^{\rm H} \right\} \right). \tag{37} \end{align} Noticing that ${\rm~E} \left\{~{\boldsymbol~w}~{\boldsymbol~w}^{\rm~H}~\right\}=~\bf{I}_N$, we have \begin{align}{\rm E}\{|x_2|^2\}=a_2^2 b_1^2\mathsf{tr} \left( \bar{\boldsymbol h}_{2}^{*} \bar{\boldsymbol h}_{2}^{\rm T} \bf{I}_N \right)=a_2^2 b_1^2 N. \tag{38} \end{align} Following the similar lines, it can be derived that \begin{align}&{\rm E}\{|x_3|^2\} = b_2^2 a_1^2 \left\|{\bar{\boldsymbol H}_\text{1} {\boldsymbol f}}\right\|^2, \tag{39} \\ &{\rm E}\{|x_4|^2\} = b_2^2 b_1^2 N, \tag{40} \\ &{\rm E}\{|x_5|^2\} = \lambda^2 b_0^2. \tag{41} \end{align} Summing over all the values yields the desired result.


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  • Figure 1

    (Color online) System model.

  •   

    Algorithm 1 Alternating optimization algorithm

    The fractional increase of the objective value is below a threshold $\varepsilon~>~0$.

    $\mathbf{Output:}$ ${\boldsymbol~\phi}^{\star}~={\boldsymbol~\phi}_{i}$

    and ${\boldsymbol~f}^{\star}={\boldsymbol~f}_{i}$.

    $\mathbf{Initialization:}$ Given feasible initial solutions ${\boldsymbol~\phi}_{0}$, ${\boldsymbol~f}_{0}$ and the iteration index $i=0$.

    repeat

    For given transmit beam ${\boldsymbol~f}_{i}$, calculate the optimal phase shift beam accoring to (14), which yields ${\boldsymbol~\phi}_{i+1}$.

    For given phase shift beam ${\boldsymbol~\phi}_{i+1}$, compute the optimal transmit beam according to (17), which yields ${\boldsymbol~f}_{i+1}$.

    $i~\leftarrow~i+1$.

    until